To find the roots of a function we need to find the solutions when y=0.
0=x^2-6x+8
⇕
x^2-6x+8 = 0
Now we have a quadratic equation. To solve it, we can use the Quadratic Formula.
ax^2+ bx+ c=0 ⇕ x=- b± sqrt(b^2-4 a c)/2 a
We first need to identify the values of a, b, and c.
0=x^2-6x+8 ⇕ 1x^2+( - 6)x+ 8=0
We can see that a= 1, b= - 6, and c= 8. Let's substitute these values into the Quadratic Formula.
We found that the solutions to the equation are x=3±1. Therefore, the roots of the given function are x=2 and x=4.
b
Similarly as in Part A, we need to write an equation first. In this case we should substitute 0 for f(x).
0=x^2-6x+9
Again, we obtained a quadratic equation. Note that the first and the last term on the right-hand side of the equation are perfect squares. This indicates that the trinomial might be a perfect square as well.
Is the first term a perfect square?
x^2=( x)^2 ✓
Is the last term a perfect square?
9= 3^2 ✓
Is the middle term twice the product of x and 3?
6x=2* x* 3 ✓
The answer to all three questions is yes! Therefore, we can write the trinomial as the square of a binomial. Note that there is a subtraction sign in the middle.
x^2-6x+9 ⇔ ( x- 3)^2
This allows us to take the square root of both sides of the equation.
Once again, we will first write the given function as an equation with y=0.
0=x^3-4x
This time we are given a polynomial equation. Since both terms contain x as a factor, let's factor it out.
0=x^3-4x ⇕ 0=x(x^2-4)
Next, we can see that the second factor is a difference of squares. Recall that it can be written as a product of two conjugate binomials.
a^2-b^2 = (a+b)(a-b)
Let's apply the above rule to the equation.
0=x(x^2-4) ⇕ 0=x(x-2)(x+2)
Now, since the right-hand side of the equation is in factored form and the left-hand side is 0, we can apply the Zero Product Property. To do so we should set each factor equal to 0.
